Hot Search : Source embeded web remote control p2p game More...
Location : Home Search - 9-1 9-2
Search - 9-1 9-2 - List

想加密自己的软件,看过一些软件的作法,思路如下:
1
、用户安装后,得出本机的机器码(硬盘的序列号,不想用网卡,因为很多机器没有网
卡)。生成机器码代码
2
、用户将该机器码EMAIL给我,我用该机器码生成一个注册码,发送给用户。注册码生成
代码
3
、用户在程序中输入该注册码,程序逆算该注册码,与机器码核对,查看是否相符。注册
码逆算、核对代码,
4
、如果用户安装后没有得到注册码,在N天后部分功能不可用。不要太容易被改(虽然说
安装后也只有少部份功能能用)。时限功能代码,


Date : 2009-01-22 Size : 44kb User : BHBH202

Check1(0).Caption = "香皂" Check1(1).Caption = "肥皂" Check1(2).Caption = "牙膏" Check1(3).Caption = "洗衣粉" Label1(0).Caption = "5.8元" Label1(1).Caption = "2.5元" Label1(2).Caption = "7.9元" Label1(3).Caption = "6.4元"-Check1 (0). Caption = "soap" Check1 (1). Caption = "soap" Check1 (2). Caption = "toothpaste" Check1 (3). Caption = "detergent" Label1 (0). Caption = "5 .8 million "Label1 (1). Caption = "2.5 yuan" Label1 (2). Caption = "7.9 yuan" Label1 (3). Caption = "6.4 yuan"
Date : 2025-12-25 Size : 15kb User : 霍永亮

单源最短路径求解-single source shortest path Solution
Date : 2025-12-25 Size : 1kb User : 郭迎风

0 / 1背包问题是一个N P-复杂问题,为了解决该问题,,将用回溯算法解决该问题。既然想选择一个对象的子集,将它们装入背包,以便获得的收益最大,则解空间应组织成子集树的形状(如图1 6 - 2所示)。该回溯算法与4 . 2节的装载问题很类似。首先形成一个递归算法,去找到可获得的最大收益。然后,对该算法加以改进,形成代码。改进后的代码可找到获得最大收益时包含在背包中的对象的集合。-0/1 knapsack problem is a P-complex issues, in order to solve the problem, and will be used backtracking algorithm to solve the problem. As to the choice of a subset of object, they will load a backpack, in order to obtain the greatest profits, the solution should be organized into space subset of the shape of the tree (Figure 1 6-2 below). The backtracking algorithms and 4. Two loading problem is very similar. Forming a recursive algorithm to get the maximum benefit available. Then, this algorithm is modified to form code. Improved code can be found at the maximum benefits included in the backpack of a collection of objects.
Date : 2025-12-25 Size : 8kb User : 周易

(1)输入字符序列,建立二叉链表。 (2)中序遍历二叉树:递归算法。 (3)中序遍历二叉树:非递归算法(最好也能实现先序,后序非递归算法)。 (4)求二叉树的高度 。 (5)求二叉树的叶子个数。 (6)建立中序线索二叉树,并实现中序遍历。 (8)借助队列实现二叉树的层次遍历。 (9)在主函数中设计一个简单的菜单,分别调试上述算法。 -(1) input sequence of characters, the establishment of binary list. (2) traversal of binary tree: recursive algorithm. (3) in order binary tree traversal: Non-recursive algorithm (the best can achieve the first order, after the sequence of non-recursive algorithm). (4) for tree height. (5) for the number of tree leaves. (6) in order to establish clues to binary tree and traversal implementation. (8) through the queue to achieve the level of binary tree traversal. (9) in the main function in the design of a simple menu, respectively, the above algorithm debugging.
Date : 2025-12-25 Size : 315kb User : jack

对弈程序采用了多种搜索算法.以下是本程序主要的类说明: 1.CEveluation类:估值类,对给定的棋盘进行估值. 2.CMoveGenerator类:走法产生器,对给定的棋盘局面搜索出所有可能的走法. 3.CSearchEngine类:搜索引擎基类. 4.CNegaMaxEngine类:负极大值法搜索引擎. 5.CAlphaBetaEngine类:采用了Alpha-Beta剪枝技术的搜索引擎. 6.CFAlphaBetaEngine类:fail-softalpha-beta搜索引擎. 7.CHistoryHeuristic类:历史启发类. 8.CAlphabeta_HHEngine类:带历史启发的Alpha-Beta搜索引擎. 9.CAspirationSearch类:渴望搜索引擎. 10.CIDAlphabetaEngine类:迭代深化搜索引擎. 11.CMTD_fEngine类:MTD(f)搜索引擎. 12.CTranspositionTable类:置换表. 13.CAlphaBeta_TTEngine类:加置换表的Alpha-Beta搜索引擎. 14.CPVS_Engine类:极小窗口搜索引擎. 15.CNegaScout_TT_HH类:使用了置换表和历史启发的NegaScout搜索引擎. 本程序还具有悔棋,还原功能,还可以记录走法.-err
Date : 2025-12-25 Size : 1.8mb User : wiali

在9格寬×9格高的大九宮格中有9個3格寬×3格高的小九宮格,並提供一定數量的數字。根據這些數字,利用邏輯和推理,在其他的空格上填入1到9的數字。每個數字在每個小九宮格內不能出現一樣的數字,每個數字在每行、每列也不能出現一樣的數字。 這種遊戲只需要邏輯思維能力,與數字運算無關。雖然玩法簡單,但數字排列方式卻千變萬化,所以不少教育者認為數獨是鍛鍊腦筋的好方法-In 9 cells wide × 9 grid of large squares in the high nine 3 grid has width × 3 grid of small squares high and provide a certain number of figures. Based on these figures, the use of logic and reasoning, in other fill in the spaces on the figures 1-9. Each number in each of the small squares should not occur within the same figures, each figure in each row, each column can not appear as figures. This game only need logical thinking ability, nothing to do with digital computing. Although the play is simple, but the figure is the arrangement of the ever-changing, many educators think Sudoku is a good way to exercise their brains
Date : 2025-12-25 Size : 156kb User : Legend_WH

Follow me超级搜索引擎V6.2(完美版)绝对是目前全国功能最优秀的超级搜索引擎之一 全国已超过1000个网站和超过100家网吧正在使用本程序 (6.3和6.2功能上变化不大,但修复了很多显示上的错误,而且优化了程序,让程序速度更快.) 主要功能: 1.本程序整合google、百度、搜狗、中搜、搜搜等等近百个搜索引擎为一体. 2.可以搜索网页、音乐、影视、软件、图片、小说、新闻、论坛、游戏、Flash、地图、购物、商业特色搜索等10多个搜索主题. 3.占用CPU资源少,支持流量大. 4.本程序版面绝对精美,所有资料来原于各大搜索引擎,完全不用维护,只要改一下网站资料和放些广告就可以了. 5.最新增加默认搜索功能,可以让用户选择百度或google或百度+google作为默认搜索设置,方便大家的使用习惯(这功能还没有其它搜索引擎能做到). 6.还增加了采用AJAX技术开发的"网址快速搜索"功能,这功能全国只有少数网站才拥有. 7.本程序已做好SEO搜索引擎优化的,可以尽快让搜索引擎收录你的网站. 8.本程序已设置好广告位,大家可以按自己的需要添加自己的广告. 9.本程序分有数据库和无数据库版本,如果对ASP不太熟的用有数据库版方便管理,对ASP很熟的话就可以使用无数据库版,无数据库版支持流量会更多. -Follow me
Date : 2025-12-25 Size : 163kb User : 王明

acm算法设计 设有n件工作分配给n个人。将工作i 分配给第j 个人所需的费用为Cij。试设计一个算法,为每一个人都分配1 件不同的工作,并使总费用达到最小。 设计一个算法,对于给定的工作费用,计算最佳工作分配方案,使总费用达到最小。 输入:第一行有1 个正整数n (1≤n≤30)。接下来的n行,每行n个数,表示工作费用。 输出:最小总费用。 例输入: 3 10 2 3 2 3 4 3 4 5 例输出:9-acm algorithm design with n pieces of work assigned to the n individuals. I will be allocated to the work of the first j for the cost of personal Cij. Try to design an algorithm for each individual a different allocation of effort and minimize the total cost. To design an algorithm for a given cost of the work to calculate the best distribution of work to minimize the total cost. Input: the first line there is a positive integer n (1 ≤ n ≤ 30). N the next line, n the number of each line, the cost of that work. Output: the minimum total cost. Cases of type: 31,023,234,345 cases output: 9
Date : 2025-12-25 Size : 347kb User : 张波

考虑由下列数所组成的表。你的工作是删去尽可能少的数使得留下来的数以升序排列 9 44 32 12 7 42 34 92 35 37 41 8 20 27 83 64 61 28 39 93 29 17 13 14 55 21 66 72 23 73 99 1 2 88 77 3 65 83 84 62 5 11 74 68 76 78 67 75 69 70 22 71 24 25 26-Considered integers consist of the following table. Your job is to delete as less as possible integers in order to let the lefts in ascending order 9 44 32 12 7 42 34 92 35 37 41 8 20 27 83 64 61 28 39 93 29 17 13 14 55 21 66 72 23 73 99 1 2 88 77 3 65 83 84 62 5 11 74 68 76 78 67 75 69 70 22 71 24 25 26
Date : 2025-12-25 Size : 35kb User :

用计算机模拟 九连环解法 还 可解任意连环-9 in a chain of computer simulation method can also be an arbitrary chain solution
Date : 2025-12-25 Size : 1kb User : 大碗

Example 编译:mpicc mst.c mst 运行:本实例中使用了5个处理器。 mpirun –np 5 mst 运行结果: Input the size of matrix:9 0 4 5 8 6 2 5 7 3 5 0 8 3 1 6 4 9 5 7 2 0 6 4 1 3 8 9 4 5 6 0 5 7 2 1 6 4 5 8 7 0 2 3 4 7 4 1 4 7 5 0 7 8 4 7 8 9 3 2 1 0 6 7 4 2 1 4 3 5 8 0 9 5 3 2 8 9 7 6 2 0 最终输出结果: the MST is: 0 0 0 0 0 2 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 1 0 1 2 0 0 0 0 0 0 0 1 0 0 1 0 0 0 2 0 0 0 2 1 1 0 2 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 2 0 0 0 0 0 0 说明:输出结果为最小生成树的边。 - Example Compile: mpicc mst.c mst Run: The example of the use of the five processors. mpirun-np 5 mst Run Results: Input the size of matrix: 9 045,862,573 508,316,495 720,641,389 456,057,216 458,702,347 414,750,784 789,321,067 421,435,809 532,897,620 The final output: the MST is: 000,002,000 000,011,000 000,001,012 000,000,010 010,002,000 211,020,100 000,001,000 001,100,000 002 million Description: The output for the minimum spanning tree edges.
Date : 2025-12-25 Size : 2kb User : aaa

加深对进程概念及进程管理各部分内容的理解;熟悉进程管理中主要数据结构的设计及进程调度算法。 1) 基本操作:进程的创建、删除。 2) 进程的基本属性: 进程名、进程时间片、进程优先级 3) 支持进程创建与撤销功能,创建后的进程在其整个生命周期中以PCB形式存在。每个进程用一个PCB表示,其内容根据具体情况设置。 进程根据其执行情况在不同队列(就绪队列、阻塞队列)间迁移。 5) 实现单处理器进程调度功能,调度等调度算法实现。 6) 管理的进程数目不少于3个,竞争使用的资源可以是CPU、内存、I/O设备等。 7) 进程可在初始时刻由系统创建,也可由其它进程在运行过程中创建。 8) 能够查看各进程状态、各进程队列内容。系统在运行过程中应能显示或打印各进程的状态及有关参数的变化情况,以便观察诸进程的运行过程及系统的管理过程。 9) 至少支持1种或者2种调度算法。-The concept of deepening of the process and process management understanding of the various parts familiar with the process management in the design of the main data structure and process scheduling algorithm. 1) basic steps: the process of creating, deleting. 2) the basic properties of the process: the process name, process time slice, the process priority 3) to support the process of creation and cancellation function, the process created throughout their life cycle in PCB form. Each process with a PCB that set the content according to specific circumstances. Process in accordance with its implementation in different queue (ready queue, blocking the queue) to another. 5) the process of single-processor scheduling, scheduling and other scheduling algorithm. 6) management of not less than three the number of processes, competitive use of resources can be CPU, memory, I/O devices. 7) The process can be created by the system at the initial time, other processes may also be c
Date : 2025-12-25 Size : 2.13mb User : moondancing

问题的提出:17世纪法国数学家加斯帕的一本《数学的游戏问题》描述了许多有趣问题,约瑟问题就是其中之一: 15名基督教徒和15名异教徒同乘一船航行,途中风浪大作,危机万分,领航者告诉大家,只要将全船的一半人投入海中,其余人就能幸免。大家都同意这个办法,并协定者30人围成一圈;由第一个人起报数,每数至第9人便把他投入海中,下一个接着从1开始报数,第9人又被投入海中,依次循环,直至剩下15人为止。问题是如何排法使投入海中的人全为异教徒? 编程的思路:此问题有人把它的答案隐示于下列詩句中: From numbers’ aid and art, (依靠数字的和魔力,) Never will fame depart! (声誉决不会离你而去!) 这句诗中元音字母依次为:oueaiaaeeiaeea,分别用1、2、3、4、5代替a、e、i、o、u,便得一排数(划线得数字表示基督教徒,不划线的表示异教徒)。 4 5 2 1 3 1 1 2 2 3 1 2 2 1 故所求排法诗4个基督教徒,5个异教徒,再2个基督教徒,1个异教徒,…可以证明,这种排法所有的基督教徒都是安全的。 -Of the problem: the 17th century French mathematician Gaspard s a "mathematical game problem " describes the many interesting questions, Joseph is one of them: 15 Christians and 15 pagans sail with the boat ride, the way storm comes up, extreme crisis, the leader tell you that as long as half of the whole ship into the sea, the rest can be spared. We all agree with this approach, and there were 30 agreements circle the first person from the reported number, the number to 9 for each person put into the sea in his next start and then reported from a number of people has been put into 9 The sea, in cycles, until the remaining 15 people so far. The question is how to arrange method to make the people into the sea are all infidels? Programming idea: the problem was the answer to it implicitly in the following verse: From numbers aid and art, (relying on numbers and magic) Never will fame depart! (reputation will never leave you!) Poem vowel sentence were: oueaiaaeeiaeea, were re
Date : 2025-12-25 Size : 12kb User : yinlingzhi

sqlite的官方文档,详细了sqlite的各种语法、数据结构等等内容 -sqlite official document, which details the various sqlite syntax, data structures and much more
Date : 2025-12-25 Size : 1.64mb User : aaf

开源 zlib 库 zlib是提供数据压缩用的函式库,由Jean-loup Gailly与Mark Adler所开发,初版0.9版在1995年5月1日发表。zlib使用DEFLATE算法,最初是为libpng函式库所写的,后来普遍为许多软件所使用。此函式库为自由软件,使用zlib授权。截至2007年3月,zlib是包含在Coverity的美国国土安全部赞助者选择继续审查的开源项目。-Is to provide open source zlib data compression library zlib library used by Jean-loup Gailly and Mark Adler of the development, first edition of version 0.9 in May 1st, 1995 publication. zlib uses DEFLATE algorithm, originally written for the libpng library, and later used by many software generally. This library is free software, use the zlib license. As of March 2007, zlib is included in Coverity' s Department of Homeland Security sponsors choose to continue the review of the open source project.
Date : 2025-12-25 Size : 639kb User : love

数据挖掘概念与技术 第二章2.9 c++ 代码-datamining concept and technology
Date : 2025-12-25 Size : 1.08mb User : 梁恒瑞

“lsa-poly” is a polynomial least square fitting tool. e.g.: lsa-poly -f data.dat -x 1 -y 2 -n “-7 -2 0 4” -oe will fit column 2 with column 1 of data.dat, and the fitting function consists of x^-7. x^-2, x^0 and x^4 terms. ===================== The source code is compressed with two precompiled binaries, the MacOS X 10.9 and Ubuntu 11 on Linux 3.0.0.14 (64 bit).
Date : 2025-12-25 Size : 85kb User : 孙谨

JAVA多代理系统开发平台(JADE)4.3.2版本(2014年9月的最新版本)的原代码.-The complete source code of JAVA Agent Development Platform (JADE), version 4.3.2 (newest at Sept 2014).
Date : 2025-12-25 Size : 17.39mb User : Lee Strong

xe=u(7);ye=u(8);the=u(9); vr=1;wr=1; delta1=0.05;delta2=0.05; k1=6;k2=6; s1=xe; s2=the+atan(vr*ye); Q=vr/(1+vr^2*ye^2); u4=(wr+Q*vr*sin(the)+k2*s2/(abs(s2)+delta2))/(1+Q*xe); w=u4; u3=ye*w+vr*cos(the)+k1*s1/(abs(s1)+delta1);
Date : 2025-12-25 Size : 11kb User : kidcad
« 12 3 4 »
CodeBus is one of the largest source code repositories on the Internet!
Contact us :
1999-2046 CodeBus All Rights Reserved.